A solution is made by dissolving 0.709 mol of nonelectrolyte solute in 815 g of

Question

A solution is made by dissolving 0.709 mol of nonelectrolyte solute in 815 g of benzene. Calculate the freezing point, Tf, and boiling point, Tb, of the solution. Constants may be found here.

(°C/m)

Normal freezing

point (°C)

(°C/m)

Normal boiling

point (°C)

Solvent Formula Kf value*

(°C/m)

Normal freezing

point (°C)

Kb value

(°C/m)

Normal boiling

point (°C)

water H2O 1.86 0.00 0.512 100.00 benzene C6H6 5.12 5.49 2.53 80.1 cyclohexane C6H12 20.8 6.59 2.92 80.7 ethanol C2H6O 1.99 –117.3 1.22 78.4 carbon
tetrachloride   CCl4 29.8 –22.9 5.03 76.8 camphor   C10H16O 37.8 176

Explanation / Answer

DTf = i*Kf*m

DTf = freezinpoint of solvent - freezinpoingt of solution = 5.49 - x

i = vanthoff factor of nonelectrolyte = 1

Kf of benzene = 5.12 c*kg/mol

m = molality of solution = (wt of solute/Mwt of solute)*(1000/wt of solvent in gr)

= 0.709/0.815 = 0.87 molal

(5.49-x) = 1*5.12*0.87

x = freezingpoint of solution Tf = 1.0356 C

DTb = i*Kb*m

DTb = boilingpoint of solution - boiling point of solvent = x-80.1

i = vanthoff factor of nonelectrolyte = 1

Kb of benzene = 2.53 c*kg/mol

m = molality of solution = (wt of solute/Mwt of solute)*(1000/wt of solvent in gr)

= 0.709/0.815 = 0.87 molal

(x-80.1) = 1*2.53*0.87

x = boiling point of solution Tb = 82.3 C

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